Answer
$\frac{dR}{dt} = 0.6$ ohms/s
Work Step by Step
Step-1: Differentiate the following equation with respect to $t$,
$$\frac{1}{R} =\frac{1}{R_1} + \frac{1}{R_2}$$
, we get,
$$-\frac{1}{R^2}\frac{dR}{dt}=-\frac{1}{R_1^2}\frac{dR_1}{dt}-\frac{1}{R_2^2}\frac{dR_2}{dt}$$
$$\implies \frac{dR}{dt}=R^2 \bigg[ \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2}\frac{dR_2}{dt} \bigg] ... (i)$$
Step-2: Find out the value of $R$ when $R_1 = 50$ ohms and $R_2=75$ ohms.
$$\frac{1}{R} = \frac{1}{50} + \frac{1}{75} \implies R = 30ohms$$
Step-3: Put $R=30$, $R_1=50$, $R_2=75$, $\frac{dR_1}{dt}=1$ & $\frac{dR_2}{dt}=1.5$ in $(i)$,
$$\frac{dR}{dt} = (30)^2 \bigg[ \frac{1}{(50)^2} \times 1 + \frac{1}{(75)^2} \times 1.5 \bigg]$$
$$\implies\frac{dR}{dt} = \frac{3}{5} = 0.6ohms/s$$
